Weak-* closed ideal of $L^\infty$ I'm having difficulty showing the following problem from a past qualifying exam.
Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space and let $I$ be a weak-* closed ideal in $L^\infty(X)$. Show that there is a set $S$ in $\Omega$ such that $I=\{f\in L^\infty: f=0 \,\text{ a.e. on } S\}$.
I'm trying to write down the definition of weak-* closed ideal and use the definitions, but could not get anywhere.
Note: by an ideal I understand a linear subspace $I$ such that $axb\in I$ whenever $a,b$ are arbitrary and $x\in I$.
 A: First of all, as $\mu$ is $\sigma$-finite, without lost of generality we may assume that $\mu$ is actually a finite measure.
Given $f\in L_\infty(X)$, writing $f^{-1}(0)$ makes sense modulo a null set, that is for two function representatives $f_1,f_2$ of $f$ the symmetric difference of $f_1^{-1}(0)$ and $f_2^{-1}(0)$ has measure 0. We shall then talk about some fixed representatives of elements of $L_\infty(X)$.
We have $f\in I$ if and only if $\mathbf{1}_{X\setminus f^{-1}(0)}\in I$. Indeed, $f\in I$, if non-zero, is invertible when restricted to $$D_n=\{x\in X\colon |f(x)|\geqslant\tfrac{1}{n}\},$$ for large enough $n$ (here invertible means invertible in the algebra $L_\infty(D_n)$). 
Moreover, $X\setminus f^{-1}(0) = \bigcup_{n=1}^\infty D_n$. Let $g_n$ be the reciprocal of $f$ on $D_n$ continued by 0 on the complement of $D_n$. Note that the sequence $(g_n f \mathbf{1}_{D_n})_{n=1}^\infty \subset I$ converges weakly* to $\mathbf{1}_{X\setminus f^{-1}(0)}$ (as the measure $\mu$ is finite), so the limit is in $I$, as it is weak* closed. Indeed, $g_n f \mathbf{1}_{D_n}= \mathbf{1}_{D_n}$. As $(\mathbf{1}_{D_n})_{n=1}^\infty$ is a bounded sequence, it is enough to test weak* convergence on a dense subset of $L_1(X)$, namely on bounded functions in $L_1(X)$. Let $g\in L_1(X)$ be a bounded function. We have
$$|\langle \mathbf{1}_{X\setminus f^{-1}(0)}-\mathbf{1}_{D_n}, g\rangle| = \Big|\int\limits_{\bigcup_{k=n+1}^\infty D_k}g\,{\rm d}\mu\Big|\leqslant \mu(\bigcup_{k=n+1}^\infty D_n)\cdot \|g\|_{L_\infty(X)}\rightarrow 0$$
as $n\to \infty$ becasue $\mu$ is a finite measure.
Then $\{\mathbf{1}_{X\setminus f^{-1}(0)}\colon f\in I\}$ is an increasing and bounded net in $L_\infty(X)$ consisting of elements in $I$ (that this set is a net follows from the fact that $I$ is closed under addition). This net converges weakly* to $$\hat{f}=\sup\{\mathbf{1}_{X\setminus f^{-1}(0)}\colon f\in I\}$$ (this supremum exists as $L_\infty(X)$ is order-complete for $\sigma$-finite measures). Thus, $\hat{f}\in I$. Consequently, $$I=\{f\in L_\infty(X)\colon f=0 \text{ on } X\setminus \hat{f}^{-1}(0)\text{ a.e.}\}.$$
